Name: Mechanism of a Chemical Reaction
Uploaded: 2019-07-19
Duration: 5 min 23 s
Description: An energy profile diagram theoretically represents a single energetic pathway for a chemical reaction. Learn the reaction mechanism with an energy diagram.

Question

The kinetics of hydrolysis of methyl acetate in excess dilute $HCl$ at $25 ° C$ were followed by withdrawing $2 mL$ of the reaction mixture at intervals of $(t)$ , adding $50 mL$ water, and titrating with baryta water. Determine the velocity constant of hydrolysis.

$t$ (in a minute)	$0$	$75$	$119$	$259$	$\infty$
Titre value (in $mL$ )	$19.24$	$24.20$	$26.60$	$32.23$	$42.03$

Accepted Answer

The hydrolysed reaction mixture of methyl acetate $({CH}_{3} {COOCH}_{3})$ is being titrated with baryta water $(Ba (OH)_{2})$ . The required volume of baryta water is given below in the table.

$t$ (in a minute) $0$ $75$ $119$ $259$ $\infty$

Titre value (in $mL$ ) $19.24$ $24.20$
$26.60$
$32.23$ $42.03$

Since, the acid-catalysed hydrolysis of methyl acetate (ester) follows pseudo-first-order kinetics with respect to ester.

${CH}_{3} {COOCH}_{3} + H_{2} O \overset{HCl}{\to} {CH}_{3} COOH + {CH}_{3} OH$

So, the rate law for the above reaction will be: $Rate = k [{CH}_{3} {COOCH}_{3}]$ .

Now as hydrolysis takes place, the volume of acetic acid increases in the reaction mixture, which will be neutralised by baryta water in the titration. So the volume of baryta water required will be directly proportional to the volume of acetic acid present in the mixture.

When the rate is studied by volume required of titre, the rate constant can be determined by the formula:

$k = \frac{2.303}{t} \log (\frac{V_{\infty} - V_{0}}{V_{\infty} - V_{t}}), where V_{0} = Volume of titre at time t = 0, V_{\infty} = Volume of titre at time t = \infty, V_{t} = Volume of titre at time t = t .$

Since we are given data at three different times, so to determine a more precise value of rate constant, we have to take average of the values of rate constant, determined at each time.

Now, at time $t = 75 \min$

$k_{75} = \frac{2.303}{75} \log (\frac{42.03 - 19.24}{42.03 - 24.20}), k_{75} = \frac{2.303}{75} \log (\frac{22.79}{17.83}), k_{75} = 3.29 \times 10^{- 3} \min^{- 1} .$

Similarly, at time $t = 119 \min$

$k_{119} = \frac{2.303}{119} \log (\frac{42.03 - 19.24}{42.03 - 26.60}), k_{119} = \frac{2.303}{119} \log (\frac{22.79}{15.43}), k_{119} = 3.29 \times 10^{- 3} \min^{- 1} .$

Similarly, at time $t = 259 \min$

$k_{259} = \frac{2.303}{259} \log (\frac{42.03 - 19.24}{42.03 - 32.23}), k_{259} = \frac{2.303}{259} \log (\frac{22.79}{9.8}), k_{259} = 3.25 \times 10^{- 3} \min^{- 1} .$

Now, taking an average of rate constants calculated at times $75, 119, 259 \min$ rate constant for the given pseudo-first-order reaction will be:

$k = \frac{k_{75} + k_{119} + k_{259}}{3} k = \frac{3.29 \times 10^{- 3} + 3.29 \times 10^{- 3} + 3.25 \times 10^{- 3}}{3} k = 3.28 \times 10^{- 3} \min^{- 1} .$

Therefore, the rate constant for the given pseudo-first-order reaction will be $3.28 \times 10^{- 3} \min^{- 1}$ .

Time (seconds)	$0$	$600$	$1200$
${KMnO}_{4}$ solution used $(mL)$	$22.8$	$13.8$	$8.2$

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